Chapter 3 Determinants, Linear Algebra 6e Lay
Chalmeta
3.1 Introduction to Determinants
Let \(A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\) and define the determinant of \(A\) as \(\det A = ad-bc\).
For any square matrix \(A\), let \(A_{ij}\) denote the submatrix formed by deleting the \(i^{\text{th}}\) row and \(j^{\text{th}}\) column of \(A\).
\(A =\begin{bmatrix}3 & 4 & -5 & -2\\ 2 & -3 & 5 & -1\\ 3 & 0 & 5 & 0 \\ 4 & 9 & 4 & 5 \end{bmatrix}\)
Find \(A_{32}\), \(A_{23}\) and \(A_{44}\)
For \(n \geq 2\) the determinant of an \(n \times n\) matrix \(A = [a_{ij}]\) is the sum of \(n\) terms of the form \(\pm a_{1j} \det A_{1j}\), with plus and minus signs alternating, where the entries \(a_{11}, a_{12}, \ldots, a_{1n}\) are from the first row of \(A\). In symbols,
\[\begin{align*} \det A &= a_{11}\det A_{11}-a_{12}\det A_{12} + a_{13}\det A_{13} - \cdots +(-1)^{1+n}a_{1n}\det A_{1n}\\ &= \sum_{j=1}^n (-1)^{1+j}a_{1j}\det A_{1j} \end{align*}\]
For a \(3 \times 3\) matrix \(A =\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}\) that would look like:
\[\det A = a \begin{vmatrix} e & f\\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h\end{vmatrix} = a(ei-hf)-b(di-fg)+c(dh-eg)\]
Find the determinant of \(A =\begin{bmatrix}3 & 4 & -5 \\ 2 & -3 & 5 \\ 3 & 0 & 5 \end{bmatrix}\)
Find the determinant of \(A =\begin{bmatrix}3 & 4 & -5 & 0\\ 2 & -3 & 5 & -1\\ 3 & 0 & 5 & 0 \\ 4 & 9 & 0 & 5 \end{bmatrix}\)
Find the determinant of \(A =\begin{bmatrix} 4 & 0 & -7 & 3 & -5\\ 0 & 0 & 2 & 0 & 0 \\ 7 & 3& -6 & 4 & -8\\ 5 & 0 & 5 & 2 & -3 \\ 0 & 0 & 9 & -1 & 2 \end{bmatrix}\)
Find the determinant of the upper triangular matrix
\[A =\begin{bmatrix} a_1 & x & x & x & x\\ 0 & a_2 & x & x & x\\ 0 & 0 & \ddots & x & x\\ \vdots & \vdots & \ddots & \ddots & x \\ 0 & 0 & \cdots & 0 & a_n \end{bmatrix}\]
Given \(n \times n\) matrix \(A = [a_{ij}]= \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}\)
Then the matrix \(A_{ij}\) is the matrix formed by deleting the \(i^{\text{th}}\) row and \(j^{\text{th}}\) column of \(A\).
The \((i,j)\)-cofactor of \(A\) is the number \(C_{ij}\) given by \[C_{ij}= (-1)^{i+j}\det A_{ij}\]
then \[\det A = a_{i1} C_{i1} + a_{i2} C_{i2} + \cdots + a_{in} C_{in}\]
This formula is called the cofactor expansion across the \(i^{\text{th}}\) row of \(A\).
A similar formula can be constructed for the cofactor expansion down the \(j^{\text{th}}\) column of \(A\): \[\det A = a_{1j} C_{1j} + a_{2j} C_{2j} + \cdots + a_{nj} C_{nj}\]
3.2 Properties of Determinants
Let \(A\) be a square matrix.
- If two rows of \(A\) are interchanged to produce \(B\), then \(\det B = -\det A\).
- If one row of \(A\) is multiplied by \(k\) to produce \(B\), then \(\det B = k \det A\).
- If a multiple of one row of \(A\) is added to another row to produce matrix \(B\), then \(\det B = \det A\).
Rule #1: If you switch two rows the sign of the determinant changes.
\[A = \begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix} \qquad B=\begin{bmatrix}3 & 4 \\ 1 & 2\end{bmatrix}\]
Rule #2: If you multiply a row of matrix \(A\) by a number, \(c\), to make matrix \(B\) then \(\det B = c \det A\).
\[A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 3 & 1 \\ 0 & 0 & -3\end{bmatrix} \qquad B=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1/3 \\ 0 & 0 & -3\end{bmatrix}\]
Rule #3: If a multiple of one row of \(A\) is added to another row to produce matrix \(B\) the determinants are the same: \(\det B = \det A\).
\[c \cdot R_i + R_j \longrightarrow R_j\]
\[A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -3 & 1 \\ 2 & 2 & -1\end{bmatrix}\]
Let \(A\) be an \(n \times n\) square matrix.
\(\det(A^{-1}) = \dfrac{1}{\det A}\)
\(\det(AB) = (\det A)(\det B)\)
\(\det(kA) = k^n(\det A)\)
\(\det A^T = \det A\)
A square matrix \(A\) is invertible if and only if \(\det A \neq 0\).
Find \(\det A\)
\[A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 5 & 6 & 7\end{bmatrix}\]
Find \(\det A\)
\[A = \begin{bmatrix} 2 & 0 & 0 & 6 \\ 1 & -7 & -5 & 0 \\ 3 & 8 & 6 & 0 \\ 0 & 7 & 5 & 4\end{bmatrix}\]
3.3 Cramer’s Rule, Volume, and Linear Transformations
Cramer’s Rule
Notation: For any \(n \times n\) matrix \(A\) and any \(b\) in \(\mathbb{R}^n\), let \(A_i(b)\) be the matrix obtained from \(A\) by replacing column \(i\) by the vector \(b\).
\[A_i(b) = [a_1~ \cdots~ a_{i-1} ~~b ~~ a_{i+1}~ \cdots ~ a_n]\]
\(A = \begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}\) then
\[A_1 \begin{pmatrix} -5 \\ -6 \\ -7\end{pmatrix} = \begin{bmatrix}-5 & 2 & 3 \\ -6 & 5 & 6 \\ -7 & 8 & 9\end{bmatrix}\]
Let \(A\) be an invertible \(n \times n\) matrix. For any \(b\) in \(\mathbb{R}^n\), the unique solution \(x\) of \(Ax=b\) has entries given by:
\[x_i = \frac{\det A_i(b)}{\det A}\]
Use Cramer’s Rule to solve the system of equations:
\[\begin{align*} 4x_1 +x_2 &=6\\ 3x_1 +2x_2 &=7 \end{align*}\]
\(A =\)
\(b =\)
\(A_1(b) =\)
\(A_2(b) =\)
\(x_1 = \dfrac{\det(A_1(b))}{\det A} =\)
\(x_2 = \dfrac{\det(A_2(b))}{\det A} =\)
Use Cramer’s Rule to determine the values of the parameter \(s\) for which the system has a unique solution and describe the solution.
\[\begin{align*} s x_1 -6s x_2 &=3\\ 3x_1 -18s x_2 &=5 \end{align*}\]
Area and Volume
Area: If \(A\) is a \(2 \times 2\) matrix, the area of the parallelogram determined by the columns of \(A\) is \(|\det A|\).
Volume: If \(A\) is a \(3 \times 3\) matrix, the volume of the parallelepiped determined by the columns of \(A\) is \(|\det A|\).
Find the area defined by the points \(A(0,-2), ~ B(5,-2), ~C(-3,1), ~D(2,1)\). (ans: 15)
Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at \((1,0,-3)\), \((1,4,4)\), and \((8,2,0)\) (ans: 82)
Linear Transformations
Let \(T: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) be the linear transformation determined by a \(2 \times 2\) matrix \(A\). If \(S\) is a parallelogram in \(\mathbb{R}^2\), then:
\[\{\text{area of } T(S)\} = |\det A| \cdot \{\text{area of } S\}\]
Let \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^3\) be the linear transformation determined by a \(3 \times 3\) matrix \(A\). If \(S\) is a parallelepiped in \(\mathbb{R}^3\), then:
\[\{\text{volume of } T(S)\} = |\det A| \cdot \{\text{volume of } S\}\]
Let \(S\) be the parallelogram determined by the vectors \(b_1 = \begin{bmatrix} -3 \\6\end{bmatrix}\) and \(b_2 = \begin{bmatrix} -3 \\ 10\end{bmatrix}\), and let \(A = \begin{bmatrix} 6 & -3 \\ -5 & 3 \end{bmatrix}\). Compute the area of the image of \(S\) under the mapping \(x \mapsto Ax\). (ans: 36)